Chemistry · 2021
JEE Main · 18 March 2021, Shift 2 · Q55
The solubility of CdSO_4 in water is 8.0 × 10^-4 mol L^-1. Its solubility in 0.01 M H_2 SO_4 solution is \ \ \ \ × 10^-6 mol L^-1. (Round off to the…
The solubility of $\displaystyle \mathrm{CdSO}_4$ in water is $\displaystyle 8.0 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$. Its solubility in $\displaystyle 0.01 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$ solution is \_\_\_\_ $\displaystyle \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$. (Round off to the Nearest Integer).
(Assume that solubility is much less than $\displaystyle 0.01$ M)
Official answer
64
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JEE Main 2021 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.